Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $x = \dfrac{r}{15r - 40} \times \dfrac{9r - 24}{3r} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $x = \dfrac{ r \times (9r - 24) } { (15r - 40) \times 3r } $ $ x = \dfrac {r \times 3(3r - 8)} {3r \times 5(3r - 8)} $ $ x = \dfrac{3r(3r - 8)}{15r(3r - 8)} $ We can cancel the $3r - 8$ so long as $3r - 8 \neq 0$ Therefore $r \neq \dfrac{8}{3}$ $x = \dfrac{3r \cancel{(3r - 8})}{15r \cancel{(3r - 8)}} = \dfrac{3r}{15r} = \dfrac{1}{5} $